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Let $G$ be a finite 2-group. Let $x$ be a non-central element of $G$ such that $C_G(x)\leq cl(x)\cup Z(G)$ where $cl(x)$ denotes the conjugacy class of $x$ in $G$. Is it true that $|C_G(x):Z(G)|=2$?

p.s.
I know that if $|G|\leq 2^8$ then the equality holds.

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    $\begingroup$ Why do you think this should be true? Why do you want it to be true, i.e. where does this question originate? $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2013 at 21:27
  • $\begingroup$ Do you have any counterexample? $\endgroup$
    – Nashenas
    Commented Aug 16, 2013 at 21:30
  • $\begingroup$ I can't speak for others, but personally I would like to know some of the motivation behind a question before spending time thinking about it $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2013 at 21:41
  • $\begingroup$ Yes, but what prompted you to check this in the first place? I assume this was not just a question that occurred to you out of the blue, so why not give us some more background? $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2013 at 21:55
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    $\begingroup$ @Carnahan The latter assertion looks incorrect. Note that this forces $C_x (G)$ to be normal in $G$. $\endgroup$
    – Amin
    Commented Aug 18, 2013 at 13:13

1 Answer 1

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So all nontrivial elements of $C_G(x)/Z(G)$ are conjugate in $G/Z(G)$.

Lemma. If $X$ is a 2-group with a subgroup $Y$ in which all nontrivial elements of $Y$ are conjugate in $X$, then $|Y| \le 2$.

Proof. Induction on $|X|$. It's clear if $X$ is abelian. Otherwise, if $|Y|>2$ then $Y \cap Z(X) = 1$, so $|YZ(X)/Z(X)| > 2$ and the inductive hypothesis applied to $X/Z(X)$ gives a contradiction.

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